Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
double degToRad(double a) { return a * acos(-1.0) / 180.0; }
double radToDeg(double a) { return a * 180.0 / acos(-1.0); }
struct point_i {
  int x, y;
  point_i() {
    x = 0;
    y = 0;
  }
  point_i(int _x, int _y) {
    x = _x;
    y = _y;
  }
};
struct point {
  double x, y;
  point() { x = y = 0.0; }
  point(double _x, double _y) : x(_x), y(_y) {}
  bool operator<(point other) const {
    if (fabs(x - other.x) > 1e-9) {
      return x < other.x;
    }
    return y < other.y;
  }
  bool operator==(point other) {
    return ((fabs(x - other.x) < 1e-9) && (fabs(y - other.y) < 1e-9));
  }
};
double dist(point p1, point p2) { return hypot(p1.x - p2.x, p1.y - p2.y); }
point rotate(point p, double theta) {
  double rad = degToRad(theta);
  return point(p.x * cos(rad) - p.y * sin(rad),
               p.x * sin(rad) + p.y * cos(rad));
}
struct line {
  double a, b, c;
};
void pointsToLine(point p1, point p2, line &l) {
  if (fabs(p1.x - p2.x) < 1e-9) {
    l.a = 1.0;
    l.b = 0.0;
    l.c = -p1.x;
  } else {
    l.a = -(double)(p1.y - p2.y) / (p1.x - p2.x);
    l.b = 1.0;
    l.c = -(double)(l.a * p1.x) - p1.y;
  }
}
int l[maxn], r[maxn];
vector<int> listPoint;
int sy, a, b;
int cariIdxPalingKanan(int n, int x, int y) {
  int kiri = 0, kanan = 2 * n - 1, mid, ans = 0;
  while (kiri <= kanan) {
    mid = (kiri + kanan) / 2;
    int titik = listPoint[mid];
    double titikDiKanan;
    point p1(x, y);
    point p2(titik, 0);
    line l;
    pointsToLine(p1, p2, l);
    titikDiKanan = -((double)sy * (double)l.b + (double)l.c) / (double)l.a;
    if (titikDiKanan < b) {
      ans = mid;
      kiri = mid + 1;
    } else {
      kanan = mid - 1;
    }
  }
  return ans;
}
int cariIdxPalingKiri(int n, int x, int y) {
  int kiri = 0, kanan = 2 * n - 1, mid, ans = 0;
  bool assigned = false;
  while (kiri <= kanan) {
    mid = (kiri + kanan) / 2;
    int titik = listPoint[mid];
    double titikDiKiri;
    point p1(x, y);
    point p2(titik, 0);
    line l;
    pointsToLine(p1, p2, l);
    titikDiKiri = -((double)sy * (double)l.b + (double)l.c) / (double)l.a;
    if (titikDiKiri > a) {
      assigned = true;
      ans = mid;
      kanan = mid - 1;
    } else {
      kiri = mid + 1;
    }
  }
  if (!assigned) {
    ans = kanan;
  }
  return ans;
}
int main() {
  int n, i, j;
  int q;
  scanf("%d %d %d", &sy, &a, &b);
  scanf("%d", &n);
  long long prefSum[2 * maxn];
  int idx = 0;
  long long sumSebelum = 0;
  for (i = 0; i < n; i++) {
    scanf("%d %d", &l[i], &r[i]);
    listPoint.push_back(l[i]);
    listPoint.push_back(r[i]);
    if (idx == 0) {
      prefSum[idx] = 0;
      prefSum[idx + 1] = prefSum[idx] + (r[i] - l[i]);
    } else {
      prefSum[idx] = prefSum[idx - 1];
      prefSum[idx + 1] = prefSum[idx] + (r[i] - l[i]);
    }
    idx += 2;
  }
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    int idxTerkanan = cariIdxPalingKanan(n, x, y);
    int idxTerkiri = cariIdxPalingKiri(n, x, y);
    if (idxTerkanan < idxTerkiri) {
      double lama;
      if (idxTerkanan % 2 == 0 && idxTerkiri % 2 == 1) {
        lama = fabs(b - a);
      } else {
        lama = 0;
      }
      printf("%.9lf\n", lama);
      continue;
    }
    int pointKiri = listPoint[idxTerkiri];
    int pointKanan = listPoint[idxTerkanan];
    point p1(x, y);
    point p2(pointKiri, 0);
    line l;
    pointsToLine(p1, p2, l);
    double titikTerkiri =
        -((double)sy * (double)l.b + (double)l.c) / (double)l.a;
    point p3(pointKanan, 0);
    pointsToLine(p1, p3, l);
    double titikTerkanan =
        -((double)sy * (double)l.b + (double)l.c) / (double)l.a;
    double pengurang = prefSum[idxTerkiri];
    double pengali = ((double)y - (double)sy) / (double)y;
    double lama = (double)(prefSum[idxTerkanan] - pengurang) * pengali;
    if (idxTerkiri % 2 == 1) {
      if (titikTerkiri >= a) {
        lama += fabs(titikTerkiri - a);
      }
    }
    if (idxTerkanan % 2 == 0) {
      if (titikTerkanan <= b) {
        lama += fabs(titikTerkanan - b);
      }
    }
    printf("%.9lf\n", lama);
  }
  return 0;
};